The proportion of $\omega$s in $A$ converges almost surely to $P(A)$ 
Let $A$ be an event in $(\Omega,\mathcal{F},P)$. We generate independent inquiries from $\Omega$ in accordance to $P$. Show that the proportion of $\omega$s in $A$ converges almost surely to $P(A)$.

My attempt:
We define a random variable $\mathbf 1_A$ and let $X_1, X_2,\ldots$ be a sequence of independent random variables distributed identically to $\mathbf 1_A$. Then $$\frac{X_1+\ldots+X_n}{n}\xrightarrow{\text{a.s}}E(\mathbf 1_A)=P(A)$$ Where the LSH represents the proportion of $\omega$s in $A$.
Would this work? Any improvements are very appreciated.
Edit: Here is what I meant: if we had $n$ experiments and in $k$ of them $\omega\in A$ then $k/n$ is the relevant share.
 A: Thanks to the comment made by Did, my confusion is somewhat cleared up now and I'll try to answer my own question.
We have $(\Omega, \mathcal F, P)$ and $A\in \mathcal F$. Then $\mathbf 1_A$ is a random variable in $(\Omega, \mathcal F)$. We know that there is another probability space, let's call it $(\Omega_\infty, \mathcal F_\infty, P_\infty)$, and there is a sequence of random variables $X_1, X_2, \ldots$ in the second space such that the sequence is independent and $P_{\mathbf 1_A}=P_{X_n}$ for the induced measures.
Since each element of $(\Omega_\infty, \mathcal F_\infty, P_\infty)$ represents a sequence of experiments on $(\Omega, \mathcal F, P)$, $X_1+\ldots+X_n$ represents the number of the $\omega$s in $A$. Furthermore, by the strong law of large numbers: $$\frac{X_1+\ldots+X_n}{n}\xrightarrow{\text{a.s}}E(X_1)$$
We can't write $\bar{X_n}\xrightarrow{\text{a.s}}E(\mathbf 1_A)$ as I did in the post because $X_1$ and $\mathbf 1_A$ aren't defined on the same space. 
But they are identically distributed, therefore $E(X_1)=E(\mathbf 1_A)$, and in turn $E(\mathbf 1_A)=P(A)$. Which is what we wanted.
I hope it makes more sense now.
